2012 AIME II Problem 1

Below is the professionally curated solution for Problem 1 of the 2012 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AIME II solutions, or check the answer key.

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Concepts:Diophantine Equationmodular arithmetic

Difficulty rating: 1870

1.

Find the number of ordered pairs of positive integer solutions (m,n)(m, n) to the equation 20m+12n=2012.20m + 12n = 2012.

Solution:

Dividing by 44 gives 5m+3n=503.5m + 3n = 503. Reducing modulo 3,3, we need 2m5032(mod3),2m \equiv 503 \equiv 2 \pmod{3}, so m1(mod3).m \equiv 1 \pmod{3}. Write m=3k+1m = 3k + 1 with k0;k \ge 0; then 3n=5035(3k+1)=49815k,3n = 503 - 5(3k + 1) = 498 - 15k, so n=1665k.n = 166 - 5k.

This is positive exactly when 5k165,5k \le 165, that is k33.k \le 33. So k=0,1,,33k = 0, 1, \ldots, 33 all work, giving 3434 ordered pairs.

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